Problem: $g(n) = 4n-2(f(n))$ $f(x) = -6x^{2}+4x$ $h(t) = 2t+3(f(t))$ $ h(f(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = -6(0^{2})+(4)(0)$ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $h(f(0))$ , which is $h(0)$ $h(0) = (2)(0)+3(f(0))$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = -6(0^{2})+(4)(0)$ $f(0) = 0$ That means $h(0) = (2)(0)+(3)(0)$ $h(0) = 0$